Electric Motor Control
10th Edition
ISBN: 9781133702818
Author: Herman
Publisher: CENGAGE L
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Chapter 1, Problem 2SQ
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A linear electrical load draws 11 A at a 0.72 lagging power factor./1 153. When a capacitor is connected, the line current dropped to 122 A and the power factor
improved to 0.98 lagging. Supply frequency is 50 Hz.
a. Let the current drawn from the source before and after introduction of the capacitor be 11 and 12
respectively. Take the source voltage as the reference and express 11 and 12 as vector
quantities in polar form.
b. Obtain the capacitor current, IC = 12 - 11, graphically as well as using complex number
manipulation. Compare the results.
c. Express the waveforms of the source current before (11(t)) and after (12(t)) introduction of the
capacitor in the form Im sin(2лft + 0). Hand sketch them on the same graph. Clearly label your
plots.
d. Analytically solve i2(t) – i1(t) using the theories of trigonometry to obtain the capacitor current
in the form, ¡C(t) = ICm sin(2πft + OC). Compare the result with the result in Part b.
Transmission line data:Data:• Active power of the load (P): 50 kW• Power factor of the load (PF): 0.8 (lagging)• Line-to-line voltage at the load (V_LC): 13.8 kV• Line resistance (R): 2 Ω• Line inductance (L): 0.8 H• Line capacitance (C): 0.0003 F• Required series compensation: 60% of the line impedance.• Line length: 250 kmDetermine:1. Characteristic impedance and propagation constant.2. The generalized long line constants A, B, C, D.3. Total voltage, current and power at the generating end.4. Voltage regulation.5. Parameters A, B, C, D of the compensation circuit.6. New generalized constants of the power system afterseries compensation.7. Conclusion of the results obtained.
3.18
In a single-phase half-wave ac-dc converter, the average value of the load
current is 1.78 A. If the converter is operated from a 240 V, 50 Hz supply
and if the average value of the output voltage is 27% of the maximum
possible value, calculate the following, assume the load to be resistive.
(a) Load resistance
(b) Firing angle
(c) Average output voltage
(d) The rms load voltage
(e) The rms load current
(f) DC power
(g) AC power
(h) Rectifier efficiency
(i) Form factor
(j) Ripple factor
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- Need a solufor number 2arrow_forwardDecompose using relationships 3 S(+) = 50 sin ³ (500πiz)arrow_forwardA linear electrical load draws I₁ A at a 0.72 lagging power factor. 11 = 153 When a capacitor is connected, the line current dropped to 122 A and the power factor improved to 0.98 lagging. Supply frequency is 50 Hz. a. Let the current drawn from the source before and after introduction of the capacitor be 1₁ and 12 respectively. Take the source voltage as the reference and express 11 and 12 as vector quantities in polar form. b. Obtain the capacitor current, Ic = I2 − I₁, graphically as well as using complex number manipulation. Compare the results. c. Express the waveforms of the source current before (i₁(t)) and after (i2(t)) introduction of the capacitor in the form Im sin(2лft + 0). Hand sketch them on the same graph. Clearly label your plots. d. Analytically solve i₂ (t) - i₁ (t) using the theories of trigonometry to obtain the capacitor current in the form, ic (t) = 1cm sin(2´ft + 0c). Compare the result with the result in Part b.arrow_forward
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